Section 12.8 #17

An Example Using Maple

>	with(plots):

Warning, the name changecoords has been redefined

>	f:=-4*x/(x^2+y^2+1);

>	fx:=diff(f,x);

>	simplify(fx);

>	fy:=diff(f,y);

>	solve({fx=0,fy=0},{x,y});

>	eval(f,{x=-1,y=0});

>	eval(f,{x=1,y=0});

The critical ordered pairs are (-1,0) and (1,0) and the corresponding critical points are (-1,0,2) and (1,0,-2).  Since applying the Second Partials Test could be rather cumbersome, it is perhaps easier to graph the surface in order to determine the nature of the critical points.

>	plot3d(f,x=-4..4,y=-4..4,axes=boxed);

Here is a different view.

>	plot3d(f,x=-4..4,y=-4..4,axes=boxed);

We can observe that (-1,0,2) is a relative maximum point and (1,0,-2) is a relative minimum point.

Below we see the Second Partials Test applied using Maple.

>	fxx:=diff(fx,x);

>	fyy:=diff(fy,y);

>	fxy:=diff(fx,y);

>	d:=fxx*fyy-(fxy)^2;

>

>	eval(fxx,{x=-1,y=0});

>	eval(d,{x=-1,y=0});

We can see analytically that since fxx(-1,0) is negative and d(-1,0) is positive the critical point (-1,0,2) is a relative maximum point based on the Second Partials Test.

>	eval(fxx,{x=1,y=0});

>	eval(d,{x=1,y=0});

We can see analytically that since fxx(1,0) is positive and d(1,0) is positive the critical point (1,0,-2) is a relative minimum point based on the Second Partials Test.

With the help of Maple the Second Partials Test is not too difficult to apply in this case.